Unilateral Shift

Abstract

If H is not separable, then it is the direct sum of separable infinite-dimensional subspaces that reduce A, and, consequently, there is no loss of generality in assuming that H is separable in the first place. In a separable Hilbert space all infinite-dimensional subspaces have the same dimension; the assertion, therefore, is just that H is the direct sum of ℵ₀ infinite-dimensional subspaces that reduce A. It is sufficient to prove the assertion for 2 in place of ℵ₀. In other words, it is sufficient to prove that for each normal operator on a separable infinite-dimensional Hilbert space there exists a reducing subspace such that both it and its orthogonal complement are infinite-dimensional. Indeed, if this is true, then there exists a reducing subspace H₁ of H such that both H₁ and H₁^⊥ are infinite-dimensional. Another application of the same result (consider the restriction of A to H₁^⊥) implies that there exists a reducing subspace H₂ of H₁^⊥ such that both H₂ and H₁^⊥ ∩ H₂^⊥ are infinite-dimensional. Proceed inductively to obtain an infinite sequence {H _n } of pairwise orthogonal infinite-dimensional reducing subspaces. If the intersection $\bigcap\nolimits_{n = 1}^\infty {{H_n}^ \bot}$ is not zero, adjoin it to, say, H₁.